no-cube.md (3040B)

1 # Solving a cube... without a cube 2 3 *In this page you can find the full text of a message I posted on the 4 [speedsolving forum](https://www.speedsolving.com/forum/threads/the-3x3x3-example-solve-thread.14345/page-273#post-1173067) 5 in 2016. It is the explanation of how I managed to write down a solution 6 for a 3x3x3 Rubik's Cube scramble using just pen an paper, without 7 a cube.* 8 9 **Scramble**: B R2 B2 R' U2 R2 D2 L B2 F2 R2 D R' U' R U' L2 R D2 B' 10 11 For this scramble, I have tried something I have had in my mind for 12 quite a long time. I have found this solution without using any cube 13 (physical nor simulator). As I would do in a blind solve, I solved edges 14 and corners separately with commutators. First, I tracked every edge 15 to see the permutation cycle (and check if there were any flipped). This 16 was easier than I had thought. I have found this. 17 18 (UR BU DR FL LB FR BD DF UF UL)(DL RB) 19 20 Now I use commutators to "solve" these cycles. Actually I "copy" these 21 cycle, so I am matching the scramble instead of solving it. No big deal, 22 I just have to invert the final solution to get one for this scramble. 23 The easiest way would be to just do a blind solve, but I struggled 24 I little bit to save moves, in order to see if I could get under the 25 80HTM limit for a legal FMC solve. I could break the first big cycle 26 into 3-cycles in any order, and I choose this one. There are probably 27 better ways to solve edges, but I didn't want to think to much about it. 28 29 (A) (UF UL UR) = [F2 U: M', U2] = F2 U R' L F2 R L' U F2 30 (B) (UF BU DR) = [M', B' R B] = R' L U' R' U R L' B' R B 31 \(C\) (UF FL LB) = [L' U L, E] = L' U L U D' B' U' B U' D 32 (D) (UF FR BD) = [U2: M, U R U'] = U2 R L' B R B' R' L U R' U 33 34 Then I solve the 2-2 cycle left. 35 36 (E) (UF DF)(DL RB) = [U B U': (L2 F2)3] = U B U' L2 F2 L2 F2 L2 F2 U B' U' 37 38 Same thing for corners. One was twisted, so I use a different notation 39 for the permutation. 40 41 UBL->BLU 42 UFL->LFD->DBR->UFR->RFD->LBD->URB->FUL 43 44 (1) (UBL UFL LFD) = [U', L D L] = U' L D L' U L D' L' 45 (2) (URB FUL UBL) = [F R' F', L] = F R' F' L F R F' L' 46 (3) (UFR RFD LBD) = [R U R', D2] = R U R' D2 R U' R' D2 47 (4) (DBR UFR URB) = [D: B2, D F2 D'] = D B2 D F2 D' B2 D F2 D2 48 49 Now I can use this commutators in any order, as long as the relative 50 order of the pieces of the same kind remains the same (i.e. I cannot use 51 (2) before (1), but I can do either (A) (B) (1), (A) (1) (B), (1) (A) 52 (B)). To cancel more moves, I used this order: 53 54 (1) (A) (2) (B) \(C\) (D) (E) (3) (4) 55 56 At this point I checked the solution on my cube and it was extremely 57 satisfactory to see everything going as I expected :) Inverting and 58 cancelling gives the following solution: 59 60 D2 F2 D' B2 D F2 D' B2 D R U R' D2 R U' R' U B U' F2 61 L2 F2 L2 F2 L2 U B' U2 R U' L' R B R' B' L R' U' D' B' 62 U B D U' L' U' L B' R' B L R' U' R U R F R' F' L' 63 F R F U' L R' F2 L' R U' F2 L D L' U' L D' L' U 64 65 alg.cubing.net 66 67 79 HTM, just under the limit :) There are many ways to improve this 68 "method", but I just wanted a success, and I am satisfied with the result.